If log _2 x + log _4 x + log _8 x + log _{16} | vedclass

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Question

(C) Given the equation: $\log _2 x + \log _4 x + \log _8 x + \log _{16} x = \frac{25}{36}$ Using the change of base formula $\log _a b = \frac{\log b}

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$\frac{1}{3}$

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(C) Given the equation: $\log _2 x + \log _4 x + \log _8 x + \log _{16} x = \frac{25}{36}$ Using the change of base formula $\log _a b = \frac{\log b}{\log a}$: $\frac{\log x}{\log 2} + \frac{\log x}{\log 4} + \frac{\log x}{\log 8} + \frac{\log x}{\log 16} = \frac{25}{36}$ Since $\log 4 = 2 \log 2$,$\log 8 = 3 \log 2$,and $\log 16 = 4 \log 2$: $\frac{\log x}{\log 2} + \frac{\log x}{2 \log 2} + \frac{\log x}{3 \log 2} + \frac{\log x}{4 \log 2} = \frac{25}{36}$ Factor out $\frac{\log x}{\log 2}$ (which is $\log _2 x$): $\log _2 x \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} ight) = \frac{25}{36}$ Calculate the sum in the bracket: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12+6+4+3}{12} = \frac{25}{12}$ So,$\log _2 x \left( \frac{25}{12} ight) = \frac{25}{36}$ $\log _2 x = \frac{25}{36} 	imes \frac{12}{25} = \frac{1}{3}$ Since $x = 2^k$,then $\log _2 x = k$. Therefore,$k = \frac{1}{3}$.

If $\log _2 x + \log _4 x + \log _8 x + \log _{16} x = \frac{25}{36}$ and $x = 2^k$,then $k$ is

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